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A322078
a(n) = n^2 * Sum_{p|n} 1/p^2, where p are primes dividing n.
11
0, 1, 1, 4, 1, 13, 1, 16, 9, 29, 1, 52, 1, 53, 34, 64, 1, 117, 1, 116, 58, 125, 1, 208, 25, 173, 81, 212, 1, 361, 1, 256, 130, 293, 74, 468, 1, 365, 178, 464, 1, 673, 1, 500, 306, 533, 1, 832, 49, 725, 298, 692, 1, 1053, 146, 848, 370, 845, 1, 1444, 1, 965, 522
OFFSET
1,4
COMMENTS
Generalized formula is f(n,m) = n^m * Sum_{p|n} 1/p^m, where f(n,0) = A001221(n) and f(n,1) = A069359(n).
LINKS
FORMULA
Sum_{k=1..n} a(k) ~ A085541 * A000330(n).
G.f.: Sum_{k>=1} x^prime(k) * (1 + x^prime(k)) / (1 - x^prime(k))^3. - Ilya Gutkovskiy, Oct 10 2019
a(n) = 1 <=> n is prime. _ Alois P. Heinz, Oct 11 2019
Dirichlet g.f.: zeta(s-2)*primezeta(s). This follows because Sum_{n>=1} a(n)/n^s = Sum_{n>=1} (n^2/n^s) Sum_{p|n} 1/p^2. Since n = p*j, rewrite the sum as Sum_{p} Sum_{j>=1} 1/(p^2*(p*j)^(s-2)) = Sum_{p} 1/p^s Sum_{j>=1} 1/j^(s-2) = zeta(s-2)*primezeta(s). The result generalizes to higher powers of p. - Michael Shamos, Mar 02 2023
EXAMPLE
a(40) = 464 because the prime factors of 40 are 2 and 5, so we have 40^2 * (1/2^2 + 1/5^2) = 464.
MAPLE
a:= n-> n^2*add(1/i[1]^2, i=ifactors(n)[2]):
seq(a(n), n=1..70); # Alois P. Heinz, Oct 11 2019
MATHEMATICA
f[p_, e_] := 1/p^2; a[n_] := If[n==1, 0, n^2*Plus@@f@@@FactorInteger[n]]; Array[a, 60] (* Amiram Eldar, Nov 26 2018 *)
PROG
(PARI) a(n) = my(f=factor(n)[, 1]~); sum(k=1, #f, n^2\f[k]^2);
(Magma) [0] cat [n^2*&+[1/p^2:p in PrimeDivisors(n)]:n in [2..70]]; // Marius A. Burtea, Oct 10 2019
CROSSREFS
Sequences of the form n^k * Sum_{p|n, p prime} 1/p^k for k = 0..10: A001221 (k=0), A069359 (k=1), this sequence (k=2), A351242 (k=3), A351244 (k=4), A351245 (k=5), A351246 (k=6), A351247 (k=7), A351248 (k=8), A351249 (k=9), A351262 (k=10).
Sequence in context: A373394 A002564 A287640 * A019428 A303547 A184753
KEYWORD
nonn
AUTHOR
Daniel Suteu, Nov 25 2018
STATUS
approved